Question

A 13.9 - g piece of metal ( specific heat capacity is 0.449 /g^ C)who whose temperature is 54.2 degrees * C was added to a sample of water at 13.4 degrees * C in a constant - pressure calorimeter of negligible heat capacity . If the final temperature of the water is 15.6 °C, calculate the mass of the water in the calorimeter .​

Answer 1

Answer:

26.2g = Mass of water in the calorimeter

Explanation:

The heat absorbed for the water is equal to the heat released for the metal. Based on the equation:

Q = m*C*ΔT

Where Q is heat, m is the mass of the sample, C is specific heat of the material and ΔT is change in temperature

Replacing we can write:

$m_{metal}*C_{metal}*dT_{metal}=m_{water}*C_{water}*dT_{water}$

13.9g * 0.449J/g°C * (54.2°C-15.6°C) = m(H₂O) * 4.184J/g°C * (15.6°C-13.4°C)

240.9J = m(H₂O) * 9.2J/g

26.2g = Mass of water in the calorimeter